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Consider the process of inserting an element into a Max Heap, where the Max Heap is represented by an array. Suppose we perform a binary search on the path from the new leaf to the root to find the position for the newly inserted element, the number of comparisons performed is
- a)θ(log2n)
- b)θ(log2 log2n)
- c)θ(n)
- d)θ(nlog2n)
Correct answer is option 'B'. Can you explain this answer?
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Consider the process of inserting an element into a Max Heap, where th...
After inserting an element at leaf in max heap we perform binary search on the path from the new leaf to the root to find the correct position for the newly inserted element. Comparisions will take θ(loglogn) but time will be θ(logn) because of swap operations.
Most Upvoted Answer
Consider the process of inserting an element into a Max Heap, where th...
O(log n)
Explanation: In a Max Heap, the parent node is always greater than or equal to its children nodes. Therefore, when inserting a new element, we compare it with its parent node and swap if necessary until it satisfies the Max Heap property.
By performing a binary search on the path from the new leaf to the root, we can quickly find the correct position for the new element based on its value compared to its parent nodes. Since a binary search works by dividing the search space in half at each step, the number of comparisons performed is proportional to the logarithm of the size of the search space, which is the height of the Max Heap.
Since the height of a Max Heap with n elements is O(log n), the number of comparisons performed when inserting a new element using binary search is also O(log n).
Explanation: In a Max Heap, the parent node is always greater than or equal to its children nodes. Therefore, when inserting a new element, we compare it with its parent node and swap if necessary until it satisfies the Max Heap property.
By performing a binary search on the path from the new leaf to the root, we can quickly find the correct position for the new element based on its value compared to its parent nodes. Since a binary search works by dividing the search space in half at each step, the number of comparisons performed is proportional to the logarithm of the size of the search space, which is the height of the Max Heap.
Since the height of a Max Heap with n elements is O(log n), the number of comparisons performed when inserting a new element using binary search is also O(log n).
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Consider the process of inserting an element into a Max Heap, where the Max Heap is represented by an array. Suppose we perform a binary search on the path from the new leaf to the root to find the position for the newly inserted element, the number of comparisons performed isa)θ(log2n)b)θ(log2 log2n)c)θ(n)d)θ(nlog2n)Correct answer is option 'B'. Can you explain this answer?
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